Defining figures on the coordinate plane by equations and inequalities. Defining figures on the coordinate plane by equations and inequalities Set of points on the coordinate plane

Let's call (x, y) ordered pair, and NS and at- the components of this pair. At the same time, it is believed that (NS 1 at 1 ) = (x 2 .y 2 ), if x 1 = x 2 and at 1 = at 2 .

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Definition 9. The set A is called the Cartesian product of the sets A and B B, whose elements are all pairs (x, y) such that x Oh, y B, i.e. BUT B = ((x, y) / x Oh, y IN).

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Let us find, for example, the Cartesian product of the sets A = (1,3} and B = (2,4,6).

BUT IN= {(1, 2);(1, 4);(1, 6);(3, 2);(3, 4);(3, 6)}.

The operation by which the Cartesian product is found is called Cartesian multiplication of sets.

Cartesian multiplication of sets possesses neither the commutative property nor the associativity property, but is associated with the operations of union and subtraction of sets by distributive properties:

for any sets A, B, C the equalities hold:

(BUT IN) C = (A WITH) (IN WITH),

(A \ B) WITH= (BUT C) \ (B WITH).

For a visual representation of the Cartesian product of number sets, a rectangular coordinate system is often used.

Let be BUT and IN - number sets. Then the elements of the Cartesian product of these sets will be ordered pairs of numbers. By depicting each pair of numbers with a point on the coordinate plane, we get a figure that will clearly represent the Cartesian product of sets BUT and IN.

We represent on the coordinate plane the Cartesian product of the sets BUT and IN, if:

a) A = {2, 6}; B ={1,4}, b) A = (2,6}; IN= , in) A =;B =.

In case a), these sets are finite and it is possible to enumerate the elements of the Cartesian product.

BUT B ={(2, 1); (2, 4); (6, 1); (6, 4)}. Let's build the coordinate axes and on the axis OH mark the elements of the set BUT, and on the axis OU - elements of the set IN. Then we represent each pair of numbers of the set АВ to points on the coordinate plane (Fig. 7). The resulting figure of four points will clearly represent the Cartesian product of these sets BUT and IN.

In case b) it is impossible to list all the elements of the Cartesian product of sets, since a bunch of IN- infinite, but you can imagine the formation process of this Cartesian product: in each pair, the first component is either 2 or 6 , and the second component is a real number from the interval .

All pairs whose first component is a number 2 and the second runs a value from 1 before 4 inclusive, are depicted by the points of the segment SD, and pairs, the first component of which is a number 6 , and the second is any real number from the interval , – segment points RS (fig. 8). Thus, in case b) the Cartesian product of the sets BUT and IN on the coordinate plane is depicted as a line segment SD and RS.

Rice. 7 Fig. 8 Fig. nine

Case c) differs from case b) in that not only the set is infinite here IN, but many BUT, so, the first component of pairs belonging to the set BUT IN, is any number in the range . Points representing the elements of the Cartesian product of sets BUT and IN, form a square SDEL (fig. 9). To emphasize that the elements of the Cartesian product are depicted by the dots of a square, it can be shaded.

test questions

Show that solving the following problems leads to the formation of a Cartesian product of sets:

a) Write down all fractions, the numerator of which is a number from the set A ={3, 4} , and the denominator is a number from the set B = (5,6, 7}.

b) Write down different two-digit numbers using numbers 1, 2, 3, 4.

Prove that for any sets A, B, C fair equality (BUT IN) С = (BUT WITH) (IN WITH). Illustrate its satisfiability for sets BUT= {2, 4, 6}, B =(1,3,5), C = (0, 1).

What figure do the points on the coordinate plane form if their coordinates are elements of the Cartesian product of sets? BUT= (- 3, 3) and IN= R

Determine which Cartesian product of which sets BUT and IN shown in Figure 10.

Rice. 10

Exercises

112. Write down all two-digit numbers whose tens digits belong to the set BUT= {1, 3, 5} , and the digits of ones - to the set B = (2,4,6).

113. Write all fractions, the numerators of which are selected from the set A = (3,5, 7}, and the denominator is from the set B ={4, 6, 8}.

114. Write all the correct fractions, the numerators of which are selected from the set A =(3, 5,7), and the denominator is from the set B = (4, 6,8}.

115. Sets are given P ={1, 2, 3}, K = (a,b}. Find All Cartesian Products of Sets R TO and K R.

116. It is known that BUT IN= ((1, 2); (3, 2); (1, 4); (3, 4); (1, 6); (3, 6)). Find out what elements the set consists of BUT and IN.

117. Write down the sets (BUT IN) WITH and BUT (IN WITH) transfer steam , if BUT=(but,b}, B = {3}, C={4, 6}

118. Make up the sets BUT B, B BUT, if:

a ) A = (a,b, s), B = (d},

b) A = { a, b}, B =  ,

in) A = (m, n,k), B = A,

G) A = { x, y, z}, B = { k, n}

119. It is known that BUT B = ((2.3), (2.5), (2.6), (3.3), (3.5), (3.6)). Find out what elements the set consists of BUT and IN.

120. Find the Cartesian product of sets A = {5, 9, 4} and IN= {7, 8, 6} and extract from it a subset of pairs in which:

a) the first component is greater than the second; b) the first component is equal to 5; c) the second component is 7.

121. List the elements belonging to the Cartesian product of sets A, B and WITH, if:

but) A = (2, 3}, B = (7, 8, 9}, WITH= {1, 0};

b) A = B= WITH= {2, 3};

in) BUT= {2, 3}, B = {7, 8, 9}, C = 

122. Draw on the coordinate plane the elements of the Cartesian product of sets A and B, if:

but) A = (x / x N,2 < NS< 4}, IN= (x / x N, x< 3};

b) A = (x / x R, 2 < х < 4}, В = {х/х  N, x< 3};

in) BUT= ; IN= .

123. All elements of the Cartesian product of two sets A and B depicted by points in a rectangular coordinate system. Write down the sets A and IN(fig. 11).

Rice. 13

124. Draw on the coordinate plane the elements of the Cartesian product of the sets X and Y, if:

but) X = (- 1.0, 1.2),Y={2, 3,4};

b) X = (- 1.0, 1.2),Y=;

in) X = [–1; 2],Y = {2, 3, 4};

G) NS= , Y = ;

e) X = [–3; 2], Y = ;

g) NS = ]–3;2[, Y= R;

h) X = (2),Y= R;

and) X =R, Y = {–3}.

125. The figures shown in fig. 14 are the result of the image on the coordinate plane of the Cartesian product of the sets X and Y. Specify these sets for each figure.

Rice. fourteen

126. Find out which Cartesian product of which two sets is depicted on the coordinate plane in the form of a half-plane. Consider all cases.

127. Establish, the Cartesian product of which two sets is depicted on the coordinate plane in the form of a right angle, which is formed at the intersection of the coordinate axes.

128. On the coordinate plane, draw a straight line parallel to the axis OH and passing through the point R(–2, 3).

129. On the coordinate plane, draw a straight line parallel to the axis OY and passing through the point R(–2, 3). Establish, the Cartesian product of which two sets is depicted on the coordinate plane in the form of this straight line.

130. On the coordinate plane, draw a strip bounded by straight lines passing through the points (–2, 0) and (2, 0) and parallel to the axis OY. Describe the set of points belonging to this strip.

131. On the coordinate plane, draw a rectangle, the vertices of which are the points BUT(–3, 5), IN(–3, 8), WITH(7, 5), D (7, 8). Describe the set of points of this rectangle.

132. Construct on the coordinate plane a set of points whose coordinates satisfy the condition:

but) NS R, at= 5;

b) NS= –3, at R;

in) NS R, | y | = 2;

G) | x| = 3, at R;

e) NS R, y≥ 4;

e) x  R, y  4;

g) NS R, | y | 4;

h) | x|  4, | y | 3 ;

and) | x | ≥1, | y | ≥ 4;

To) | x | ≥ 2, y R.

133. On the coordinate plane, depict the elements of the Cartesian product of sets X and Y, if:

but) X = R, Y = {3}; b) X = R, Y = [–3; 3]; in) X = .

134. On the coordinate plane, draw a figure F if

but) F= ((x, y)| x = 2, y R}

b) F= ((x, y) |x R, y = –3);

in) F= ((x, y) | x 2, y R};

G) F= ((x, y) | x TO,y≥ – 3};

e) F= ((x, y) | | x | = 2, y R};

e) F= ((x, y) | x R, | y | = 3).

135. Construct a rectangle with vertices at points (–3, 4), (–3, –3), (1, –3), (1, 4). Specify the characteristic property of points belonging to this rectangle.

136. On the coordinate plane, draw straight lines parallel to the OX axis and passing through the points (2, 3) and (2, –1). Establish which Cartesian product of which two sets is depicted on the coordinate plane in the form of a strip enclosed between the constructed straight lines.

137. On the coordinate plane, draw straight lines parallel to the OY axis and passing through the points (2, 3) and (–2, 3). Establish which Cartesian product of which two sets is depicted on the coordinate plane in the form of a strip enclosed between the constructed straight lines.

138. Draw in a rectangular coordinate system the set X Y, if:

a) X = R; Y ={ y at R, |at| < 3},

b) NS= {x/ x R, |NS| > 2}; Y= (y / y R, |at| > 4}.

On the topic of this chapter, the student should be able to:

Set sets in different ways;

Establish relationships between sets and depict them using Euler-Venn diagrams;

Prove the equality of two sets;

Perform operations on sets and illustrate them geometrically using Euler-Venn diagrams;

Split the set into classes using one or more properties; evaluate the correctness of the classification performed.

Let it be given equation in two variables F (x; y)... You have already seen how to solve such equations analytically. Many solutions to such equations can be represented in the form of a graph.

The graph of the equation F (x; y) is the set of points of the coordinate plane xOy, the coordinates of which satisfy the equation.

To plot a two-variable equation, you first express the y variable in the equation in terms of the x variable.

Surely you already know how to build various graphs of equations with two variables: ax + b = c - line, yx = k - hyperbola, (x - a) 2 + (y - b) 2 = R 2 - circle whose radius is equal to R, and the center is at point O (a; b).

Example 1.

Plot the equation x 2 - 9y 2 = 0.

Solution.

Factor the left side of the equation.

(x - 3y) (x + 3y) = 0, that is, y = x / 3 or y = -x / 3.

Answer: Figure 1.

A special place is occupied by the assignment of figures on a plane by equations containing the sign of an absolute value, on which we will dwell in detail. Consider the stages of building graphs of equations of the form | y | = f (x) and | y | = | f (x) |.

The first equation is equivalent to the system

(f (x) ≥ 0,
(y = f (x) or y = -f (x).

That is, its graph consists of graphs of two functions: y = f (x) and y = -f (x), where f (x) ≥ 0.

To plot the second equation, two functions are plotted: y = f (x) and y = -f (x).

Example 2.

Plot equation | y | = 2 + x.

Solution.

The given equation is equivalent to the system

(x + 2 ≥ 0,
(y = x + 2 or y = -x - 2.

We build a set of points.

Answer: Figure 2.

Example 3.

Plot the equation | y - x | = 1.

Solution.

If y ≥ x, then y = x + 1, if y ≤ x, then y = x - 1.

Answer: Figure 3.

When plotting graphs of equations containing a variable under the modulus sign, it is convenient and rational to use area method, based on dividing the coordinate plane into parts in which each submodule expression retains its sign.

Example 4.

Plot the equation x + | x | + y + | y ​​| = 2.

Solution.

In this example, the sign of each submodule expression depends on the coordinate quarter.

1) In the first coordinate quarter, x ≥ 0 and y ≥ 0. After expanding the module given equation will look like:

2x + 2y = 2, and after simplification x + y = 1.

2) In the second quarter, where x< 0, а y ≥ 0, уравнение будет иметь вид: 0 + 2y = 2 или y = 1.

3) In the third quarter x< 0, y < 0 будем иметь: x – x + y – y = 2. Перепишем этот результат в виде уравнения 0 · x + 0 · y = 2.

4) In the fourth quarter, for x ≥ 0, and y< 0 получим, что x = 1.

We will plot this equation in quarters.

Answer: Figure 4.

Example 5.

Draw the set of points whose coordinates satisfy the equality | x - 1 | + | y ​​- 1 | = 1.

Solution.

The zeros of the submodule expressions x = 1 and y = 1 divide the coordinate plane into four regions. Let's expand the modules by area. Let's arrange it in the form of a table.

Answer: Figure 5.

On the coordinate plane, figures can be specified and inequalities.

Inequality graph with two variables is the set of all points of the coordinate plane, the coordinates of which are solutions of this inequality.

Consider algorithm for constructing a model of solutions to an inequality with two variables:

1. Write down the equation corresponding to the inequality.
2. Plot the equation from step 1.
3. Select an arbitrary point in one of the half-planes. Check if the coordinates of the selected point satisfy the given inequality.
4. Graphically depict the set of all solutions to the inequality.

Consider, first of all, the inequality ax + bx + c> 0. The equation ax + bx + c = 0 defines a line that divides the plane into two half-planes. In each of them, the function f (x) = ax + bx + c preserves its sign. To determine this sign, it is enough to take any point belonging to the half-plane and calculate the value of the function at this point. If the sign of the function coincides with the sign of the inequality, then this half-plane will be the solution to the inequality.

Let's consider examples of graphical solutions to the most common inequalities in two variables.

1) ax + bx + c ≥ 0. Figure 6.

2) | x | ≤ a, a> 0. Figure 7.

3) x 2 + y 2 ≤ a, a> 0. Figure 8.

4) y ≥ x 2. Figure 9.

5) xy ≤ 1. Figure 10.

If you have questions or want to practice depicting on the model plane the sets of all solutions to two-variable inequalities using mathematical modeling you can spend free 25-minute lesson with online tutor after you sign up. For further work with the teacher, you will have the opportunity to choose a tariff plan that suits you.

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It is often necessary to depict on the coordinate plane a set of solutions to an inequality in two variables. A solution to an inequality with two variables is a pair of values ​​of these variables, which turns the given inequality into a true numerical inequality.

2y+ 3x< 6.

First, let's build a straight line. For this, we write the inequality in the form of the equation 2y+ Zx = 6 and express y. Thus, we get: y = (6-3x) / 2.

This straight line splits the set of all points of the coordinate plane into points located above it and points located below it.

Take a meme from each area by control point, for example A (1; 1) and B (1; 3)

The coordinates of point A satisfy this inequality 2y + Zx< 6, т. е. 2 . 1 + 3 . 1 < 6.

Point B coordinates not satisfy this inequality 2 ∙ 3 ​​+ 3 ∙ 1< 6.

Since this inequality can change sign on the straight line 2y + Zx = 6, the inequality is satisfied by the set of points of the area where point A is located. Shade this area.

Thus, we have depicted a set of solutions to the inequality 2y + Zx< 6.

Example

We represent the set of solutions to the inequality x 2 + 2x + y 2 - 4y + 1> 0 on the coordinate plane.

Let us first construct a graph of the equation x 2 + 2x + y 2 - 4y + 1 = 0. We select the equation of the circle in this equation: (x 2 + 2x + 1) + (y 2 - 4y + 4) = 4, or (x + 1) 2 + (y - 2) 2 = 2 2.

This is the equation of a circle with center at point 0 (-1; 2) and radius R = 2. Construct this circle.

Since this inequality is strict and the points lying on the circle itself do not satisfy the inequality, we construct the circle with a dotted line.

It is easy to check that the coordinates of the center O of the circle do not satisfy this inequality. The expression x 2 + 2x + y 2 - 4y + 1 changes its sign on the constructed circle. Then the inequality is satisfied by points located outside the circle. These points are shaded.

Example

We represent on the coordinate plane the set of solutions of the inequality

(y - x 2) (y - x - 3)< 0.

First, let us plot the graph of the equation (y - x 2) (y - x - 3) = 0. It is the parabola y = x 2 and the straight line y = x + 3. Let's build these lines and note that the change in the sign of the expression (y - x 2) (y - x - 3) occurs only on these lines. For point A (0; 5), we define the sign of this expression: (5-3)> 0 (that is, this inequality is not satisfied). Now it is easy to mark the set of points for which this inequality is satisfied (these areas are shaded).

An Algorithm for Solving Inequalities in Two Variables

1. Let us reduce the inequality to the form f (x; y)< 0 (f (х; у) >0; f (x; y) ≤ 0; f (x; y) ≥ 0;)

2. We write down the equality f (x; y) = 0

3. Recognize the graphs written on the left side.

4. We build these graphs. If the inequality is strict (f (x; y)< 0 или f (х; у) >0), then - by strokes, if the inequality is not strict (f (x; y) ≤ 0 or f (x; y) ≥ 0), then - by a solid line.

5. Determine how many parts of the graphics the coordinate plane was divided into

6. Select a control point in one of these parts. Determine the sign of the expression f (x; y)

7. Arrange signs in other parts of the plane, taking into account alternation (as in the method of intervals)

8. We select the parts we need in accordance with the inequality sign that we solve, and apply shading